I stumbled upon this number theory problem while I was solving another problem. Here is the equation: $$3^kn + 3^{k-1} + 2^m(3^{k-1} + 2h) = 2^{m+l}n$$ where $k \geq 3, h,l,m,n\in\mathbb{N}$, $n$ is odd and $n$ is not a multiple of $3$. My impression is that it does not have a solution. However, I have not progressed on the problem anymore than that. Could you please help?


  • $\begingroup$ I removed the tag exponential sum because that's about the $e$ number. $\endgroup$ – quanta Apr 15 '11 at 23:03

I think you can use infinite descent to show that the equation does not have a solution. We know that $n= 3k+1$ for $k$ an odd integer or $n =3k+2$ for $k$ an even integer. If $(k,h,l,m,n)$ is a solution, then $2^{m+1}$ divides the LHS and $n$ divides the LHS.

  • $\begingroup$ it is a good suggestion but I do not yet see a way out since the coefficients of $n$ are not similar powers. $\endgroup$ – Chulumba Mar 16 '11 at 18:12
  • 1
    $\begingroup$ This general form of infinite descent is equivalent to mathematical induction (it is simply a contrapositive reformulation of it). So without any specific details, it is not saying much to speculate that induction might play some role. It does for almost all analogous problems - whether explicitly, or implicitly in various lemmas invoked. $\endgroup$ – Bill Dubuque Mar 16 '11 at 18:39
  • $\begingroup$ @Bill, that is a good point. $\endgroup$ – Chulumba Mar 16 '11 at 19:14
  • $\begingroup$ Thanks for the link on infinite descent. Great outline. $\endgroup$ – John Jun 13 '13 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.